I am having some problem determining all the cases for finding the conjugate of a function.
Find the conjugate of:
(i) $f(x) = e^x$ , $x ∈ \mathbb {R}$
For this, I determined that $f^* = x^*ln(x^*) – x^*$ and I thought this was the answer but I had to consider other cases which I am not sure how to determine.
(ii) $f(x) = \rho ||x||_1$
For this, I have to consider the cases when $||z||_\infty \le \rho$ and $||z||_\infty \gt \rho$ but I don't know what is the process to think of these cases. Any help is appreciated
Best Answer
This really is a case of grinding through.
(i) $f^*(\alpha) = \sup_x \alpha x - e^x$. Note that for $\alpha <0$, if $x<0$ we have $\alpha x -e^x > \alpha x -1$ and so $f^*(\alpha) = \infty$. If $\alpha = 0$, then choose large negative $x$ we see that $f^*(0) = 0$. Finally, as you have done in the question, for $\alpha>0$ we have $f^*(\alpha) = \alpha( \log \alpha -1)$.
Hence $f^*(\alpha) = \begin{cases}\infty, & \alpha < 0 \\0, & \alpha =0 \\ \alpha( \log \alpha -1), & \alpha >0 \end {cases}$.
(ii) $f^*(\alpha) = \sup_x \sum_k \alpha_k x_k - \rho |x_k|$.
It is straightforward to see that if $\| \alpha\|_\infty > \rho$ then $f^*(\alpha) =\infty$.
If $\| \alpha\|_\infty \le \rho$, then $f^*(\alpha) \le \sup_x \sum_k (|\alpha_k| - \rho )|x_k| \le 0$, so $f^*(\alpha) = 0$.
Hence $f^*(\alpha) = \begin{cases} 0,& \|\alpha\|_\infty \le \rho \\ \infty, & \|\alpha\|_\infty > \rho\end {cases}$.